If a=cos left(frac{8 pi}{11}right)+i sin left | vedclass

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(B) Given $a = \cos \left(\frac{8 \pi}{11}\right) + i \sin \left(\frac{8 \pi}{11}\right) = e^{i \frac{8 \pi}{11}}$.Since $a^{11} = e^{i 8 \pi} = 1$,$a

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$-\frac{1}{2}$

Vedclass · Answer

(B) Given $a = \cos \left(\frac{8 \pi}{11}\right) + i \sin \left(\frac{8 \pi}{11}\right) = e^{i \frac{8 \pi}{11}}$.Since $a^{11} = e^{i 8 \pi} = 1$,$a$ is an $11^{th}$ root of unity.The sum of all $11^{th}$ roots of unity is $1 + a + a^2 + \dots + a^{10} = 0$.Thus,$a + a^2 + a^3 + a^4 + a^5 + a^6 + a^7 + a^8 + a^9 + a^{10} = -1$.Since $a^{11} = 1$,we have $a^{10} = \bar{a}$,$a^9 = \bar{a^2}$,$a^8 = \bar{a^3}$,$a^7 = \bar{a^4}$,and $a^6 = \bar{a^5}$.Substituting these into the sum,we get $(a + \bar{a}) + (a^2 + \bar{a^2}) + (a^3 + \bar{a^3}) + (a^4 + \bar{a^4}) + (a^5 + \bar{a^5}) = -1$.Using the property $z + \bar{z} = 2 \operatorname{Re}(z)$,we have $2 \operatorname{Re}(a) + 2 \operatorname{Re}(a^2) + 2 \operatorname{Re}(a^3) + 2 \operatorname{Re}(a^4) + 2 \operatorname{Re}(a^5) = -1$.Therefore,$2 \operatorname{Re}(a + a^2 + a^3 + a^4 + a^5) = -1$,which implies $\operatorname{Re}(a + a^2 + a^3 + a^4 + a^5) = -\frac{1}{2}$.

If $a=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right)$,then $\operatorname{Re}\left(a+a^2+a^3+a^4+a^5\right)=$

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